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Unformatted text preview: PHYS 132, WEEK 7, CH 16 16.9. Solve: Because the atomic mass number of gold is 197, 1.0 mole of gold has a mass of 197 g or 0.197 kg. The volume of 1.0 mole of gold is V= L = 3 M rAu (0.197 kg ) ^ fiL= 3 ~ 19, 300 kg m 1 3 = 2.17 cm 16.12. Solve: Let TF = TC = T: TF = 9 TC + 32 fi T = 5 9 5 T + 32 fi T = 40 That is, the Fahrenheit and the Celsius scales give the same numerical value at 40 . 16.19. Model: Treat the gas in the sealed container as an ideal gas. Solve: (a) From the ideal gas law equation pV = nRT, the volume V of the container is V= nRT p = (2.0 mol )(8.31 J mol K )[(273 + 30) K] 1.013 10 5 Pa = 0.0497 m 3 Note that pressure must be in Pa in the ideal gas law. (b) The beforeandafter relationship of an ideal gas in a sealed container (constant volume) is p1 V T1 = p2V T2 fi p2 = p1 T2 T1 = (1.0 atm ) (273 + 130 ) K = 1.33 atm (273 + 30) K Note that gaslaw calculations must use T in kelvins. 16.24. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve: (a) The number of moles of oxygen is n= M M mol = 50 g 32 g mol 23 1 = 1.563 mol 23 (b) The number of molecules is N = nN A = (1.563 mol ) 6.02 10 2 2 ( mol 23 ) = 9.41 10 2 3 . (c) The volume of the cylinder V = p r L = p (0.10 m ) ( 0.40 m ) = 1.257 10 N V nRT V m . Thus, 25 = 9.41 10 1.257 10 2 m 3 = 7.49 10 m 3 (d) From the idealgas law pV = nRT we can calculate the absolute pressure to be p= = (1.563 mol )(8.31 J mol K )( 293 K ) 1. 257 10 2 m 3 = 303 kPa where we used T = 20 C = 293 K. But a pressure gauge reads gauge pressure: p g = p  1 atm = 303 kPa  101 kPa = 202 kPa 16.31. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 106 m3, and T1 = 20C = 293 K. This produces a pressure nRT1 (0.1 mol)(8.31 J / mol K) (293 K) 6 p1 = = = 4.87 10 Pa = 12.02 atm V1 50 10 6 Pa An ideal gas process obeys p2V2/T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure p2 = T2 V1 T1 V 2 p1 = 1 200 50 12.02 atm = 48.1 atm (b) 16.39. Model: Assume the gas in the solar corona is an ideal gas. Solve: The number density of particles in the solar corona is N V . Using the idealgas equation, pV = Nk B T fi N p = V k BT ( 0.03 Pa ) N = = 1.1 1015 particles m 3 23 6 V 1.38 10 J K 2 10 K ( )( ) 16.45. Model: Assume that the steam (as water vapor) is an ideal gas. Solve: The volume of the liquid water is V= m = nM mol = r r 20 1.013 10 Pa 10,000 10 m (0.018 kg mol ) pV ^ M mol = RT r (8.31 J mol K )(473 K) 1000 kg m 3 ( 5 )( 6 3 ( ) ) = 9.28 105 m3 = 92.8 cm3 16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant. Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the beforeandafter relationship of an ideal gas for an isochoric (constant volume) process, p1 T1 = p2 T2 fi p2 = T2 T1 p1 = 273 + 45 ^ ( 44.7 psi ) = 49 .4 psi 273 + 15 Your tire gauge will read a gauge pressure p g = 49.4 psi  14.7 psi = 34.7 psi. 16.54. Model: Assume that the gas is an ideal gas. Solve: Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is proportional to 1/V1 for isothermal processes. 16.56. Model: Assume that the helium gas is an ideal gas. Visualize: Please refer to Figure P16.56. Process 1 2 is isochoric, process 2 3 is isothermal, and process 3 1 is isobaric. Solve: The number of moles of helium is M 8.0 g n= = = 2.0 mol M mol 4 g mol Using the idealgas equation, V1 = nRT1 p1 = (2.0 mol )(8.31 J mol K )[( 273 + 37) K] 2 1.013 10 5 Pa ( ) = 0.0254 m3 For the isochoric process V2 = V1, and p1 T1 = p2 T2 fi p 2 = p1 657 + 273 ^ = (2 atm ) = 6 atm 37 + 273 T1 6 atm T2 For the isothermal process, the equation p3V3 = p2V2 is V3 = V2 p2 p3 = 0.0254 m ( 3 ~ ) 2 atm ^ = 0.0762 m 3 For the isothermal process, T3 = T2 = 657 C. 16.61. Model: Assume CO2 gas is an ideal gas. Solve: (a) The molar mass for CO2 is M mol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes 0.227 mol of gas at 0 C. With V1 = 10,000 cm3 = 0.010 m3 and T1 = 0 C = 273 K, the pressure is p1 = nRT1 V1 = (0.2273 mol )(8.31 J mol K )(273 K ) 0.010 m 3 = 5.156 104 Pa = 0.509 atm (b) From the isothermal compression, p 2 V2 = p 1 V1 fi V2 = V1 p1 p2 = 0.010 m ( 3 ) 0.509 atm ^ 3 3 3 = 1.70 10 m = 1700 cm 3.0 atm From the isobaric compression, T3 = T2 V3 1000 cm ^ = ( 273 K ) ~ = 161 K = 112 C V2 1700 cm 3 3 (c) 16.69. Model: Assume that the compressed air is an ideal gas. Visualize: Please refer to Figure CP16.69. Solve: (a) Because the piston is floating in equilibrium, Fnet = (p 1 patmos)A w = 0 N where the piston's crosssectional area A = p r = p (0.050 m ) = 7.854 10 m and the piston's weight w = (50 kg)(9.8) = 490 N. Thus, w 490 N 5 p1 = + p atmos = + 1.013 10 Pa = 1.637 105 Pa A 7.854 10 3 m 2 Using the idealgas equation p1V1 = nRT1, (1.637 10 Pa)Ah 1 = ( 0.12 mol)(8.31 J / mol K)[( 273 + 30 ) K] 5 ( ) 2 2 3 2 With the value of A given above, this equation yields h1 = 0.235 m = 23.5 cm. (b) When the temperature is increased from T1 = 303 K to T2 = (303 + 100) K = 403 K, the volume changes from V1 = Ah1 to V2 = Ah2 at a constant pressure p2 = p1. From the beforeandafter relationship of the ideal gas: p1 ( Ah1 ) p ( Ah 2 ) = 2 T1 T2 fi h2 = p1 T2 p2 T1 403 K ^ h1 = (1) ~ (0.235 m ) = 0.313 m = 31.3 cm 303 K Thus, the piston moves h2 h1 = 7.8 cm. ...
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This note was uploaded on 05/05/2008 for the course PHYS 132 taught by Professor Sharpe during the Spring '08 term at Cal Poly.
 Spring '08
 Sharpe
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